Does this series have a closed-form representation?

Question

The following sum represents the number of relevant kinds of lines in an N-dimensional tic-tac-toe game, which is why I am interested in finding a closed form, but it also is the sum of all possible combinations of N unique elements when any number of the elements from 1 to N can be chosen, which is also cool, and seems like the kind of thing that would have an elegant transcendental form involving factorials and stuff.

$$ S = \sum_{j=1}^{N} {N! \over j!(N-j)!}, N \in \mathbb{Z}_{+} $$

So is there an easy way to find a closed form here?

Answer 1

It is well-known that $$\sum_{j=0}^N\frac{N!}{j!(N-j)!}=2^N$$ Your series $S$ is missing the first term, hence $$S=2^N-1$$

Answer 2

$$\binom{N}{j}=\frac{N!}{j!(N-j)!}.$$

$$(1+x)^N=\sum_{j=0}^N\binom{N}{j}x^j.$$

$$S=(1+1)^N-\binom{N}{0}=2^N-1$$

Answer 3

$S = 2^{N}-1$. You can get this by noticing that $\sum\limits_{i=0}^{N}\binom{N}{i} = (1+1)^{N} = 2^{N}$ by the binomial theorem.